@infinitesimal I am aware of the distinction between "a-thing-in-itself" and the symbol we use to represent it, certainly $9$ and $IX$ refer to some quality common to both descriptions
@TheArtist The Archangel of integrals came to me one night and handed me the Book of Knowledge with all Integrals, series and limits a man can think of. That was the start ... (I'm just kidding ...:-))
Neurobiologists seem to be of the opinion lately that there is a certain region of the brain with a highly-localized and specific function for distinguishing quantity
@Chris'ssis This Archangel-he, or she? What did they look like?
@TheArtist I began with a textbook one of my cousines didn't use anymore since she gave up the high school, it was a book ready for thrash bin. I took it, studied it and then I began to like more and more the stuff in it, especially integrals, series and limits. The story is far more complex than that, of course.
@DavidWheeler The most impressive of all was the Basel series I found at one of the pages. There was no proof nowhere, so I was thinking I might find one at that time.
However, I couldn't imagine how one can possibly get that very strage $\pi^2/6$. It looked to me like some sort of magical thing more than something scientifical.
That series marked me profoundly. I remember that I was skimming book after book in a library hoping to find that series with a proof and found anything for a while. At that time I didn't have internet, so less possibilities for information.
I didn't have rest until I found that magical explanation (one of them, involving Fourier series) behind the very beautiful result.
For a long while I was asking questions and I was the only one to answer them. That's the disadvantage of being self-educated, there is no one to give you a hand when needed.
I said many things on MSE, this is true though, although it sounds ridiculuosly crazy: starting from a book that was prepared to be thrown at trash bin. :-)
Hello! I have a permutation group that is central symmetric after the group operation has been applied a certain number of times. How could I go about proving this?
@Committingtoachallenge I suggest you just focus on your class work and forget about the extra books. Also there is no point following my list of books, hehe.
@DavidWheeler, do you know offhand the cardinality of the Hilbert Cube? In particular, is $\displaystyle |\mathbb{R}| = \left| \prod_{k=1}^\infty [0, 1] \right|$ ?
They both have the same characteristic polynomial: $x^3$. It's clear the second one has minimal polynomial $x^3$ as well. What is the minimal polynomial of the first one?
@Committingtoachallenge What I'm saying is, there is NO WAY, if we started with the first matrix, we would wind up with the 2nd under ANY similarity transform.
Okay I must not understand something, I just wasn't sure if we were allowed to take $C={J_1}_{2\times 2} \bigoplus {J_2}_{1\times 1}$ for $$ C = \left( \begin{array}{cc|c} 0 & 1 & 0 \\ 0 & 0 & 0 \\ \hline 0 & 0 & 0 \end{array} \right). $$
@Committingtoachallenge And I'm trying to convince you we CAN, because if we were forced to have the second matrix as the first matrix's Jordan form, we get a contradiction.
@MaryStar yes I would say so. While it is possible to have functions where this is false under relatively weak conditions partial derivatives commute, and this can be assumed. See this wiki page
I think this course is amazing, just too time consuming. But maybe it needs to be so we learn everything(or are exposed to everything) within the 13 week teaching period
Here is one more thing you can do-you can make the assumption, and see if you obtain a solution. Then check that your solution satisfies the problem. In that case, you can verify that $u \in C^{2,2}$. This is often done in "real life" applications.
@Committingtoachallenge I would argue this way, by (complete) induction on the size of the matrix-reduce it to JNF. If $A^n = I$, then $J^n = I$. Now we can write: $J = \begin{bmatrix}J_1&0\\0&J_2\end{bmatrix}$. Prove $J^n = \begin{bmatrix}J_1^n&0\\0&J_2^n\end{bmatrix}$
If $\mathscr{P}$ is a prime ideal and $\mathscr{A, B} $ usual ideals, then is it true that $(\mathscr{PA}) \cap \mathscr{B}=\mathscr{P}(\mathscr{A} \cap \mathscr{B})$?
And, in general, does anyone know of a ring where the ideals are particularly pathological, and not too hard to manipulate?
It would be really nice to have a counter-examples ring on hand.
The point of using induction is to avoid even discussing the super-diagonal, for 1x1 matrices, these are all we have, and if it's true (that the Jordan blocks are all diagonal) for any k x k matrix, where k < m, breaking the JNF into two blocks, gives us two square matrices of size less than m, so we can apply our induction hypothesis.
@DavidW I know I didn't use induction on that(but I did fix induction for what I originally thought you were referring to), but I got rid of my $1\times 1$ argument, since the roots must be distinct. Can you see if it works please? meta.math.stackexchange.com/a/4726/142198
I am trying to understand correspondance between binary trees with n+1 leaves and dyck word od length 2n from this section of the wikipedia page of binary trees. Can somebody explain the last paragraph to me?
@Committingtoachallenge Lol, well you used induction, but we don't need it to prove $A^n = PJ^nP^{-1}$. But that part's OK, and we can use it to show that $J^n = I$.
Hi. I wonder if someone could help me check my answer for an exercise: Assume that 28 % of voters favored party A at some point. A later opinion poll gave a result of 30 % of voters favoring part A. What is the minimum sample size allowing us to discard the null hypothesis of no change in voter preference?
Treating the poll as a binomial test approximated by the normal distribution, I arrived at n>= 1373.
7 Chat guidelines | $\LaTeX$ in chat | MSE chat dwellers: pin your location (just for fun) [instructions] - mar 10 at 19:30 by Committing to a challenge
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$
$\dfrac{dy}{dt}=x+y=t+y$ , we have $y=y_0e^t-t-1=y_0e^x-x-1$
$\dfrac{du}{dt}=0$ , letting $u(0)=F(y_0)$ , we have $u(x,y)=F(y_0)=F((x+y+1)e^{-x})$
$u(x,1-...
In my lecture notes there is the following example on which we have applied the method of characteristics:
$$u_t+2xu_x=x+u, x \in \mathbb{R}, t>0 \\ u(x,0)=1+x^2, x \in \mathbb{R}$$
$$$$
$$(x(0), t(0))=(x_0,0)$$
We will find a curve $(x(s), t(s)), s \in \mathbb{R}$ such that $\sigma '(s)=...
The other line of reasoning we were following (which was completely different) is that $x^n - 1$ has $n$ distinct roots. So no eigenvalue can be taken more than once.
@ABeautifulMind I'm not sure if there's a way to fix that, and here's why: the way to make a link is [text](url), so the extra square brackets around the text can't be interpreted correctly. The system interprets only the text between the first left and right brackets, that's why the right bracket is black on the pinned message you're referring to
Here is something to test to see if you understand. The matrix $\begin{bmatrix}-\frac{1}{2}&-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}&-\frac{1}{2}\end{bmatrix}$ satisfies $A^3 = I$. Diagonalize it.
Hey @DanielFischer!!! Could you maybe take a look at this? http://math.stackexchange.com/questions/1190644/what-initial-value-do-i-have-to-take-at-the-beginning/1190678#1190678 Could we pick $(x(0),y(0))=(x_0,1-x_0)$ ?
@ABeautifulMind To fix it, someone would either unpin Committing to a Challenge's comment and pin a new one, or go back in the transcript and fix the error there. The link still works, so why bother?
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$
$\dfrac{dy}{dt}=x+y=t+y$ , we have $y=y_0e^t-t-1=y_0e^x-x-1$
$\dfrac{du}{dt}=0$ , letting $u(0)=F(y_0)$ , we have $u(x,y)=F(y_0)=F((x+y+1)e^{-x})$
$u(x,1-...
In my lecture notes there is the following example on which we have applied the method of characteristics:
$$u_t+2xu_x=x+u, x \in \mathbb{R}, t>0 \\ u(x,0)=1+x^2, x \in \mathbb{R}$$
$$$$
$$(x(0), t(0))=(x_0,0)$$
We will find a curve $(x(s), t(s)), s \in \mathbb{R}$ such that $\sigma '(s)=...
